2617 to the nearest thousand
2 answers:
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Given:
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).
Answer: at the values where cos(x) = 0
Justification:
1) tan(x) = sin(x) / cos(x).
2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.
3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.
Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.
You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.
Justification:
1) tan(x) = sin(x) / cos(x).
2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.
3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.
Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.
You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.